The iPhone six has been out for few years now and a lot of data has been collect

Question

The iPhone six has been out for few years now and a lot of data has been collected. A marketing firm wants to model the price (p) of an iPhone six and Weekly Demand (s). Below is a table of data that have been collected Price = p, (s) 150 170 190 210 230 250 | weekly Demand = s, (l ,000s) 217 206 195 191 176 175 Round answers to 4 decimal places. a) Find the correlation coefficient, be careful with the sign b) Perform a hypothesis test to see if the correlation is statistically significant. What is the p-value? c) Is the correlation statistically significant at the 0.01 significance level? Select an answer d) Find the linear model that best fits this data using regression and enter the model below. Be careful what letter(s) you use. Preview c What does the model predict will be the weekly demand if the price of an iPhone six is $213? thousand d) According to the model at what should the price be set in order to have a weekly demand of 205,500 iPhone sixes? Hint: Set weekly demand at 205.5 and solve for price. Round your answer to the nearest dollar

Explanation / Answer

Solution:

Run below r code:

price <- c(150,170,190,210,230,250)
weeklydemand <- c(217,206,195,191,176,175)
cor.test(price,weeklydemand)

output:

Pearson's product-moment correlation

data: price and weeklydemand
t = -10.82, df = 4, p-value = 0.0004139
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.9982544 -0.8505896
sample estimates:
cor
-0.9833434

r=-0.9833

Solutionb:

p=0.0004

Solutionc:

cor.test(price,weeklydemand,conf.level=0.99)

output:

Pearson's product-moment correlation

data: price and weeklydemand

t = -10.82, df = 4, p-value = 0.0004139

alternative hypothesis: true correlation is not equal to 0

99 percent confidence interval:

-0.9991424 -0.7176155

sample estimates:

cor

-0.9833434

p=0.0004

p<0.01

SIGNIFICANT

ANSWER:YES

Solutiond:

rmod3.lm <- lm(weeklydemand~price)
coefficients(rmod3.lm)

(Intercept) price
280.1904762 -0.4342857

Regression eq is

weekly demand=280.1905-0.4343(price)

Solutione:

price=$213

weekly demand=280.1905-0.4343(213)

weekly demand =187.6846

weekly demand=188(rounding to nearest integer)

Solutionf:

weekly demand=280.1905-0.4343(price)

Given weekly demand=205.5

205.5=280.1905-0.4343(price)

0.4343(price)=280.1905-205.5

price= 74.6905/0.4343

price=171.979

price=172 Dollars

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